Integral test for convergence?

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It's precisely the fact that it has infinitely-many negative terms; the Integral Test only works when the sequence $a_n$ we're taking the series of is the restriction of a positive non-increasing integrable function $f(x)$. To understand why, it is useful to understand the general idea of the proof of the integral test. At this site is a good illustration of the Riemann sums being taken and how they actually relate to the infinite series (this is one of the nice examples when the proof is essentially the picture).

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Updated on October 23, 2020

Comments

  • dfg
    dfg about 3 years

    Why does the series' terms have to be non-negative to use the integral test?

    Consider the series:

    $$\sum_{n = 1}^{\infty}\frac{n\cos n - \sin n}{n^2}$$

    Even though it has negative terms, why can't the integral test be used with the integral being:

    $$\int_1^\infty\frac{x\cos x - \sin x}{x^2}$$

    Since the series can be interpreted as the Riemann sum of the function above?

    • Daniel Fischer
      Daniel Fischer over 9 years
      You need a monotonic function to be sure that you can estimate the partial sums. If the function oscillates, how are $f(n)$ and $\int_n^{n+1} f(t)\,dt$ related?
    • Thomas Andrews
      Thomas Andrews over 9 years
      HInt: You need more than it being non-negative. It needs to be decreasing.
    • mjb4
      mjb4 over 9 years
      @Thomas Andrews But if it is just decreasing then the value of the integral is not equal to the sum or?
    • dfg
      dfg over 9 years
      @DanielFischer But you can have monotonic series with negative terms can't you? (I know my example wasn't monotonic) If so, why would the terms have to be positive?
    • Daniel Fischer
      Daniel Fischer over 9 years
      @dfg You can use the integral comparison test for a non-negative non-increasing function, or a non-positive non-decreasing function. These are symmetric, so one usually only mentions the non-negative case.
    • dfg
      dfg over 9 years
      @DanielFischer Thanks for the help.
    • Thomas Andrews
      Thomas Andrews over 9 years
      @mjb4 Nobody said the integral is equal to the sum.
    • mjb4
      mjb4 over 9 years
      @thomasAndrews yeah I know! just wanted to make clear!
  • Pedro
    Pedro over 9 years
    The last equality is true, but that's not how the integral test works.
  • mjb4
    mjb4 over 9 years
    yeah but if your series is positiv and the integral is finite than the sum converges too or?
  • Marcel Besixdouze
    Marcel Besixdouze over 9 years
    This would actually be an example where the conclusion of the integral test holds, although the hypotheses don't. Both the series from 1 to infinity and the integral from 1 to infinity of $\cos(\pi x)$ fail to exist.